In a triangle ABC,DE is parallel to BC and D is the midpoint of side AB.Find the perimeter of triangle ABC when AE=4.5cm,DE=5cm and DB=3.5 cm

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Please find below the solution to the asked query:

From given information we form our diagram , As :

Here , BC | | DE and D is mid point of AB , So

DB  =  AD = 3.5 cm   (  As given DB  = 3.5 cm )

From converse of mid point theorem we get

$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{DE}} = \frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \frac{\mathrm{AD}}{\mathrm{DE}} = \frac{\mathrm{AD} + \mathrm{DB}}{\mathrm{BC}} \\ \Rightarrow \frac{3.5}{5} = \frac{3.5 + 3.5}{\mathrm{BC}} \\ \Rightarrow \frac{3.5}{5} = \frac{7}{\mathrm{BC}} \\ \Rightarrow \frac{1}{5} = \frac{2}{\mathrm{BC}} \\ \mathbf{\Rightarrow }\mathbf{BC}\mathbf{=}\mathbf{ }\mathbf{10} \end{gathered}$$
And if D is mid point of AB and DE | | BC , so from converse of mid point theorem we get that E is mid point of AC , So

AE  =  CE  =  4.5 cm  (  As given AE  = 4.5 cm )

Then,

AC =  AE + CE  = 4.5 + 4.5 = 9 cm

Therefore,

Perimeter of triangle ABC  =  AB  + BC  + AC  = 10 + 7 + 9 =  26 cm                                                           ( Ans )


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